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BY 4.0 license Open Access Published by De Gruyter August 22, 2020

Liouville property of fractional Lane-Emden equation in general unbounded domain

  • Ying Wang EMAIL logo and Yuanhong Wei

Abstract

Our purpose of this paper is to consider Liouville property for the fractional Lane-Emden equation

(Δ)αu=upinΩ,u=0inRNΩ,

where α ∈ (0, 1), N ≥ 1, p > 0 and Ω ⊂ ℝN–1 × [0, +∞) is an unbounded domain satisfying that Ωt := {x′ ∈ ℝN–1 : (x′, t) ∈ Ω} with t ≥ 0 has increasing monotonicity, that is, ΩtΩt for t′ ≥ t. The shape of Ω := limt→∞ Ωt in ℝN–1 plays an important role to obtain the nonexistence of positive solutions for the fractional Lane-Emden equation.

MSC 2010: 35J60; 35B53

1 Introduction

In this paper, we consider Liouville property for the fractional Lane-Emden equation

(Δ)αu=upinΩ,u=0inRNΩ, (1.1)

where α ∈ (0, 1), p > 0, Ω is an unbounded domain in ℝN with N ≥ 1, and (–Δ)α with α ∈ (0, 1) is the fractional Laplacian defined in the principle value sense,

(Δ)αu(x)=cN,αlimϵ0+RNBϵ(0)u(x)u(x+z)|z|N+2αdz,

here Bϵ(0) is the ball with radius ϵ centered at the origin and cN,α > 0 is the normalized constant. We say that u is a bounded solution of (1.1) if uC(ℝN) ∩ L(ℝN) and u satisfies (1.1) pointwisely.

As an important property, the Liouville theorem for Lane-Emden equation has attracted a lot of attentions by many mathematician by the application in the derivation of uniform bound via blowing-up analysis. Note that the nonexistence of stable solution is studied in [1] by finite morse index with restrictions on the boundary and at infinity. Without the zero Dirichlet boundary condition, Liouville results could be obtained by Hadamard property in [2, 3], by iterating the decaying rate at infinity in [4] and by Hardy estimates in [5].

It is known that the Leray-Schauder degree theory is a very useful method for deriving solutions of elliptic equations on bounded domains. The essential step is to obtain a uniform bound by considering a sequence of solutions {un}n such that un(xn) = ∥unL = Mn → +∞ as n → +∞ for some {xn}nΩ. Then let Ωn = {x ∈ ℝN : 1Mnκ x + xnΩ} and vn(x)=1Mnun(1Mnκx+xn) for some κ > 0, then {vn}n is uniformly bounded, and the limit v of {vn}n is a solution of related limit semilinear equation in the limit domain Ω. Note that for C2 domain Ω, the limit domain of Ωn is either Ω = ℝN or Ω = ℝN–1 × (0, ∞). While if the original domain contains a cone point on the boundary, then the limit domain has the third possibility that Ω is a cone. As a consequence, the nonexistence of positive solutions to the limit equation in a cone has to be involved additionally. As far as we know, the nonexistence of elliptic equation depends on the shape of limit domain at infinity in some direction. Our concern in this article is to consider the non-existence of elliptic equations in one type of unbounded domain.

Recently, qualitative properties of solutions for nonlocal elliptic equations have been studied extensively, such as the existence of weak solutions or very weak solutions in [6, 7] by variational methods, a survey in [8] on variational methods, large solution [9] by Perron’s method, the regularities in [10], Pohozaev’s identity [11] and Liouville results [12, 13, 14]. In [15] it develops the method of moving plane in fractional setting to obtain the classification of critical elliptic equations in an integral form and then it is used to obtain nonexistence of bounded solutions for semilinear elliptic equations in half space [12, 13, 16] subject to various boundary type conditions. In particular, Fall and Weth in [13] obtained the nonexistence of positive bounded solution of (1.1) when Ω=R+Nandp<N1+2αN12α, by applying the method of moving plane, via some interesting estimates of Green’s kernel in R+N . It is worth noting that star-shaped domain with respect to infinity is involved in obtaining the nonexistence of fractional Lane-Emden equation in [14] and it is an important notation in our derivation of nonexistence to (1.1).

Before stating our main result, we introduce the following notations.

(𝓓) Given 𝓞 ⊂ ℝM × [0, +∞) and t ∈ [0, +∞), with M ≥ 1 being an integer, denote

Ot:={xRM:(x,t)O}.

For t ≥ 0, 𝓞t is nonempty, bounded and the mapping t ↦ 𝓞t is increasing in the following sense

OtOtif0tt<+.

Denote

O=t[0,)Ot. (1.2)

Now we introduce two types domains:

  1. star-shaped domain 𝓞 ⊂ ℝM, if there exists e ∈ 𝓞 such that for every point x 𝓞, the segment {te + (1 – t)(xe): t ∈ [0, 1]} is contained in 𝓞;

  2. star-shaped domain 𝓞 ⊂ ℝM with respect to infinity, if there exists a point e ∈ ℝM ∖ 𝓞̄ such that for every point x ∈ 𝓞, the half-line {e + t(xe): t ≥ 1} is contained in 𝓞.

The main results state as follows.

Theorem 1.1

Assume that α ∈ (0, 1), Ω ⊂ ℝN–1 × [0, +∞) is a C0,1 domain verifying (𝓓) and Ω is given as (1.2) with 𝓞 = Ω. Then problem (1.1) has no nonnegative, nontrivial and bounded solutions if one of the following holds:

  1. Ω = ℝN–1 and 0 < p < N1+2αN12α if N > 1 + 2α, otherwise for p > 0, 1 ≤ N ≤ 1 + 2α;

  2. Ω ⊂ ℝN–1 is a star-shaped domain with respect to infinity and 1 ≤ p N1+2αN12α with N > 1 + 2α;

  3. Ω = ℝN–2 × [0, +∞) and 1 < p < N2+2αN22α for N > 2 + 2α, otherwise for p > 1, 1 ≤ N ≤ 2 + 2α;

  4. Ω is bounded, C2 star-shape inN–1 and N > 1 + 2α, p N1+2αN12α .

Our basic tool is the traditional method of moving plane, involved by [17] in the fractional setting, we develop this traditional method of moving planes to obtain the increasing monotonicity in the direction xN and reduce problem (1.1) into

(Δ)RN1αu=upinΩ, (1.3)

subject to zero Dirichlet boundary condition when Ω ≠ ℝN–1, where (Δ)RN1α is the fractional Laplacian in ℝN–1. Then the nonexistence results could be obtained for (1.3) as in [13, 14].

Remark 1.1

In the particular case that Ω is a cone such as {x = (x′, xN) ∈ ℝN : xN > θ|x′|} for some θ > 0, problem (1.1) has no nonnegative, nontrivial and bounded solutions.

If Ω also verifies the similar assumptions of Theorem 1.1, we can repeat our above procedure and derive the following corollary directly:

Corollary 1.1

Under the assumptions of Theorem 1.1, we assume that the domain Ω ⊂ ℝN–2 × [0, +∞) is a C2 domain verifying (𝓓) and Ω∞,∞ is given by (1.2) with 𝓞 = Ω∞,∞. Then problem (1.1) does not admit nonnegative, nontrivial and bounded solutions if one of the following holds:

  1. Ω∞,∞ = ℝN–2 and 0 < p < N2+2αN22α if N > 2 + 2α, otherwise for p > 0, 1 ≤ N ≤ 1 + 2α;

  2. Ω∞,∞ ⊂ ℝN–2 is a star-shaped domain with respect to infinity and 1 ≤ p N2+2αN22α with N > 2 + 2α;

  3. Ω∞,∞ = ℝN–3 × [0, +∞) and 1 < p < N3+2αN32α for N > 3 + 2α, otherwise for p > 1, 1 ≤ N ≤ 3 + 2α;

  4. Ω∞,∞ is bounded, star-shape C2 domain inN–2 and N > 2 + 2α, p N2+2αN22α .

2 The proof of nonexistence results

For the domain Ω verifying (𝓓), we shall prove that the solution of (1.1) has the xN-increasing property by using the method of moving planes. To this end, we introduce the following notations. For λ > 0, denote

Σλ={x=(x,xN)RNΩ|xN<λ}, (2.1)
Tλ={x=(x,xN)RN|xN=λ}, (2.2)
uλ(x)=u(xλ)andwλ(x)=uλ(x)u(x), (2.3)

where xλ = (x′, 2λxN) for x = (x′, xN) ∈ ℝN. For any subset A of ℝN, we write Aλ = {xλ : xA} the reflection of A with regard to Tλ. Since Ωt is bounded for t ≥ 0, then the domain Σλ is always bounded for any λ > 0.

Proposition 2.1

Under the hypotheses of Theorem 1.1, let u be a nonnegative nontrivial solution of (1.1), then

u(x,t)u(x,s)forst,xRN1. (2.4)

Proof

We divide the proof into two steps.

  1. By the assumption of Ω R+N , we may assume

    λ0=inf{λ0:wλ0inΣλ}.

    The purpose of this step is to show that if λ > λ0 is close to λ0, then wλ > 0 in Σλ. To this end, let Σλ = {xΣλ | wλ(x) < 0}, we first claim that if λ > λ0 is close to λ0, then

    Σλ=. (2.5)

    By contradiction, we assume (2.5) is not true, that is Σλ ≠ ∅. We denote

    wλ+(x)=wλ(x),xΣλ,wλ+(x)=0,xRNΣλ, (2.6)
    wλ(x)=0,xΣλ,wλ(x)=wλ(x),xRNΣλ (2.7)

    It is obvious that wλ+(x)=wλ(x)wλ(x) for all x ∈ ℝN. By direct computation, for x Σλ , we have

    (Δ)αwλ(x)=(ΩΩλ)(ΩλΩ)wλ(z)|xz|N+2αdz(ΣλΣλ)(ΣλΣλ)λwλ(z)|xz|N+2αdz(Σλ)λwλ(z)|xz|N+2αdz=I1I2I3.

    We look at each of these integrals separately. Since u = 0 in ΩλΩ and uλ = 0 in ΩΩλ, then

    I1=ΩλΩuλ(z)|xz|N+2αdzΩΩλu(z)|xz|N+2αdz=ΩλΩuλ(z)(1|xz|N+2α1|xzλ|N+2α)dz0,

    by the fact that uλ ≥ 0 and |xzλ| > |xz| for all x Σλ and zΩλΩ.

    In order to fix the sign of I2, note that wλ(zλ) = –wλ(z) for any z ∈ ℝN and then

    I2=ΣλΣλwλ(z)|xz|N+2αdz+ΣλΣλwλ(zλ)|xzλ|N+2αdz=ΣλΣλwλ(z)(1|xz|N+2α1|xzλ|N+2α)dz0,

    since wλ ≥ 0 in Σλ Σλ and |xzλ| > |xz| for all x Σλ and zΣλ Σλ . Finally, since wλ(z) < 0 for z Σλ , we have

    I3=Σλwλ(zλ)|xzλ|N+2αdz=Σλwλ(z)|xzλ|N+2αdz0.

    Hence, we obtain that for all λ > λ0,

    (Δ)αwλ(x)0,xΣλ (2.8)

    and then for x Σλ ,

    (Δ)αwλ+(x)(Δ)αwλ(x)=(Δ)αuλ(x)(Δ)αu(x). (2.9)

    Combining (1.1) with (2.9) and (2.6), we have that

    (Δ)αwλ+(x)(Δ)αuλ(x)(Δ)αu(x)=uλp(x)up(x)=φ(x)wλ+(x),xΣλ,

    where

    φ(x)=uλp(x)up(x)uλ(x)u(x),|φ|2p[uλp1+up1]2puLp1.

    Hence, we have

    Δαwλ+(x)φ(x)wλ+(x),xΣλ (2.10)

    and observe that wλ+ = 0 in (Σλ)c, then we have that

    supxΣλwλ+(x)cR(Σλ)2αφwλ+L(Σλ)cR(Σλ)2αφL(Σλ)supxΣλwλ+(x),

    that is,

    1c2pR(Σλ)2αuLp1, (2.11)

    where c is a positive constant independent of Σλ and

    R(Σλ)=infr>0:|Br(x)Σλ|12|Br(x)|,xΣλ.

    Choosing λ > λ0 close enough to λ0, R( Σλ ) is small, a contraction is derived from (2.11) and then

    wλ=wλ+0inΣλ.

    But this is a contradiction with the definition of Σλ , so we have that wλ ≥ 0 in Σλ.

    We next claim that if wλ ≥ 0 and wλ ≢ 0 in Σλ, then wλ > 0 in Σλ. Assuming this claim is true, we complete the proof, since the function u is positive in Ω and u = 0 on ∂Ω, so that wλ is positive on ∂Ω∂Σλ and then, by continuity wλ ≢ 0 in Σλ.

    Now we prove above claim. In fact, assume there exists x0Σλ such that wλ(x0) = 0, that is, uλ(x0) = u(x0). One hand we have that

    (Δ)αwλ(x0)=(Δ)αuλ(x0)(Δ)αu(x0)=uλp(x0)up(x0)=0. (2.12)

    On the other hand, let Aλ = {(x1, x′) ∈ ℝN | x1 > λ}, since wλ(zλ) = –wλ(z) for any z ∈ ℝN and wλ(x0) = 0, we find

    (Δ)αwλ(x0)=Aλwλ(z)|x0z|N+2αdzAλwλ(zλ)|x0zλ|N+2αdz=Aλwλ(z)(1|x0z|N+2α1|x0zλ|N+2α)dz.

    Since |x0zλ| > |x0z| for zAλ, wλ(z) ≥ 0 and wλ(z) ≢ 0 in Aλ, then

    (Δ)αwλ(x0)<0, (2.13)

    which contradicts (2.12).

  2. Our purpose of this step is to move the planes forward for any λ > λ0 up to

    λ1:=sup{λ>λ0|wμ>0inΣμforanyμ(λ0,λ)}=+.

    By contradiction, we assume that λ1 < +∞. Since for any λ ∈ (0, λ1), we have that wλ > 0 in Σλ, by the continuity, we derive that wλ1 ≥ 0 in Σλ1 and wλ1 ≢ 0 in Σλ1. Thus, by the claim just proved above, we have wλ1 > 0 in Σλ1.

    Now we claim that if wλ > 0 in Σλ1, then there exists ϵ > 0 such that wλ > 0 in Σλ for all λ ∈ [λ1, λ1 + ϵ). This claim directly implies that λ1 = +∞, completing Step 2.

    In fact, if λ1 < +∞, then under our hypotheses, we have that Σλ is compact. Denote

    Dμ={xΣλ|dist(x,Σλ)μ}

    for μ > 0 small. Since wλ1 > 0 in Σλ1 and Dμ is compact, then there exists μ0 > 0 such that wλ1μ0 in Dμ. By continuity of wλ(x), for ϵ > 0 small enough and any λ ∈ (λ1, λ1 + ϵ), we have that

    wλ0inDμ.

    As a consequence,

    ΣλϵΣλϵDμ

    and R( Σλϵ ) is small if ϵ and μ are small. Using (2.8) and proceeding as in Step 1, we have that for all x Σλϵ ,

    (Δ)αwλϵ+(x)=(Δ)αuλϵ(x)(Δ)αu(x)(Δ)αwλϵ(x)(Δ)αuλϵ(x)(Δ)αu(x)=uλϵp(x)up(x)=φ(x)wλϵ+(x),

    where φ(x)=uλϵp(x)up(x)uλϵ(x)u(x) is bounded. Since wλϵ+ = 0 in (Σλϵ)c and | Σλϵ | is small, for ϵ and μ small, from the analysis Step 1, we obtain that wλϵ ≥ 0 in Σλϵ. Since λϵ > 0 and wλϵ ≢ 0 in Σλϵ, as before we have wλϵ > 0 in Σλϵ, completing the proof of the claim. The proof ends.□

Proof of Theorem 1.1

We prove this argument by contradiction. Assume that u ≥ 0 is nontrivial solution of (1.1). When Ω verifies (𝓓), then by Proposition 2.1, we have that u satisfies (2.4). Let

um(x)=u(x,xN+m),xRN,

then {um}m is an increasing and bounded sequence of functions and satisfies

(Δ)αu=upinΩm.

Note that ΩtΩt for 0 ≤ tt′ < +∞,

ΩmΩkifk>m,limm+Ωm=RN,

from the regularity results in [9, Theorem 2.1] and the stability property [9, Theorem 2.4], we obtain that

u:=limm+um

is a bounded classical solution to

(Δ)αu=upinΩ×R,u=0inRN(Ω×R)ifΩRN1.

By the increasing property of {um}m, we have that

u(x,xN)=u(x,yN)foranyxN,yNR,

which implies that u is xN-independent. Letting v(x′) = u(x′, xN), by the standard argument, we have that

(Δ)RN1αv(x)=cN(Δ)αu(x),xΩ×R,

then v is a positive, bounded and classical solution for

(Δ)RN1αv=cNvpinΩ, (2.14)

subject to v = 0 in ℝN–1Ω if Ω ≠ ℝN–1.

A contradiction is derived by the fact that problem (2.14) has no positive bounded classical solution when

  1. Ω = ℝN–1 and 0 < p < N1+2αN12α if N > 1 + 2α, otherwise for p > 0, 1 ≤ N ≤ 1 + 2α by [13, Theorem 1.2];

  2. Ω ⊂ ℝN–1 is a star-shaped domain with respect to infinity and 1 ≤ p N1+2αN12α with N > 1 + 2α by [14, Theorem 1.5];

  3. Ω = ℝN–2 × [0, +∞) and 1 < p < N2+2αN22α for N > 2 + 2α, otherwise for p > 1, 1 ≤ N ≤ 2 + 2α by [14, Theorem 1.2];

  4. Ω is bounded, C2 and star-shape in ℝN–1, N > 1 + 2α, and p N1+2αN12α . by [14, Corollary 1.2].

As a consequence, we obtain the nonexistence results of (1.1) in Theorem 1.1.□

Proof of Corollary 1.1

When problem (1.1) has a positive solution u, then Proposition 2.1 and regularity results guarantee that v(x′) = limm→+∞u(x′, xN + m) is a positive, bounded and classical solution for problem

(Δ)RN1αv=cNvpinΩandv=0inRN1Ω. (2.15)

Furthermore, we note that v∞,∞(x″) = limm→+∞ v(x″, xN–1 + m) is a positive, bounded and classical solution of

(Δ)RN2αv,=cNvpinΩ,andv,=0inRN2Ω,, (2.16)

subject to v∞,∞ = 0 in ℝN–2Ω∞,∞ if Ω∞,∞ ≠ℝN–2. Then the same conclusion is obtained as the proof of Theorem 1.1.□

Acknowledgements

Y. Wang is supported by NNSF of China, No: 11661045, by the Jiangxi Provincial Natural Science Foundation, No: 20202ACBL201001, 20202BAB201005, and by Key R&D plan of Jiangxi Province, No:20181ACE50029. Y. Wei is supported by NNSF of China, No: 11871242.

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Received: 2019-04-26
Accepted: 2020-07-08
Published Online: 2020-08-22

© 2021 Ying Wang and Yuanhong Wei, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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